Product of positive and negative integers is negative
An Introduction to Real Analysis
2025-10-26
Show that if \(m < 0\) and \(n > 0\), then \(mn < 0\).
If \(m < 0\) and \(n > 0\), then \(0 - m, n - 0 \in \symbb{N}\), so \((0 - m) \cdot (n - 0) \in \symbb{N}\). Hence \[\begin{align*} (0 - m) \cdot (n - 0) &= (0 + (-m)) \cdot (n + (-0))\\[0.75em] &= (-m) \cdot (n + 0)\\[0.75em] &= (-m) \cdot n\\[0.75em] &= ((-1) \cdot m) \cdot n\\[0.75em] &= (-1) \cdot (m \cdot n)\\[0.75em] &= -mn\\[0.75em] &= -mn + 0\\[0.75em] &\in \symbb{N}. \end{align*}\] This implies that \(-mn > 0\). By Exercise 1.2.5 from Katznelson and Katznelson (2024), it then follows that \[\begin{align*} -(-mn) &= (-1) \cdot ((-1) \cdot (mn))\\[0.75em] &= ((-1) \cdot (-1)) \cdot (mn)\\[0.75em] &= 1 \cdot mn\\[0.75em] &= mn\\[0.75em] &< 0. \end{align*}\]